A (5,3),B(3,-2) are two fixed points; find the equation to the equation to the locus of a point P which moves so that the area of the triangle PAB is 9 uints.

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Solution

$Let P(h,k) be any point on the locus,then, A r e a (P A B) = 9 s q . u n i t s \Rightarrow \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) | = 9 \Rightarrow | 5 (- 2 - k) + 3 (k - 3) + h (3 + 2) | = 18 \Rightarrow | - 10 - 5 k + 3 k - 9 + 5 h | = 18 \Rightarrow | 5 h - 2 k - 19 | = 18 \Rightarrow 5 h - 2 k - 19 \pm 18 \Rightarrow 5 h - 2 k - 19 \pm 18 = 0 \Rightarrow 5 h - 2 k - 37 = 0 o r 5 h - 2 k - 1 = 0 H e n c e, t h e l o c u s o f (h, k) i s 5 x - 2 y - 37 = 0 o r, 5 x - 2 y - 1 = 0$

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